Array counting instances of negative and numbers that go beyond array size

[luv2program]

Hello i have a problem with array counting instances of negative and numbers that go beyond array size. I made an array to count the number of instances of only numbers that are not beyond the size limit of 30. Been smashing my head for a few hours. Anything help is much appreciated!

currently my code only runs from 1-30 inputs.

[Kevin C]

// bubble sort algorithm
for( int i = 0; i <= input; i++)

it should be < instead of <= (but this will not cause an error)

// extracting number of instances for each element in array
for( int i = 0; i <= input; i++)

it should be < instead of <=

int count[30]
count[arr[i]]++;

this wont work unless all the input numbers are smaller than 30

// printing array of elements with number of instances
for( int i = 0; i < input; i++)

it should be: i<input-1

Your main problem is the size of the 'count' array. It must be as large as the greatest number that will be input by the user, plus 1.

[Bdanielz]

currently my code only runs from 1-30 inputs.

sorry more my fault than yours. What do you want your code to do that it is not?

[Kevin C]

Perhaps you should do it without the 'count' array. Just set int cnt=0, and then go through array 'arr' to count equal elements, increasing cnt by 1 on each iteration. WHen you find an element not equal to the previous element, print out cnt and reset it back to 0.

Also, it is better to use std::vector than c-style arrays.

[luv2program]

@Kevin C

how would i count instances of negative numbers as well?

@Bdanielz

i want it to count the instances of negative numbers and the greatest number of the user input and print them.

[Kevin C]

The solution with cnt will work for negative numbers, too. I have edited the previous answer, so reread it please.

[luv2program]

@Kevin C
thank you for your input and will try to work on it and will get back to u

[Kevin C]

That inner loop is unnecessary. Just one loop. And you should be doinng the last part (printing out and count=0) only when arr[i]!=arr[i+1].

- And: in for loop, i < input-1
- and dont forget count++ when you remove the inner loop.
- use:
cout << arr[i] << "repeats: " << count << endl;

Edit: and I left out another thing for the end... you can figure it out

[Kevin C]

Have you tried using a debugger? It will be easier. Are you using an IDE and which one?

[luv2program]

.

[Kevin C]

Nah, your solution is still not good enough. And you have made it too complicated.

Try to figure out this one:

	// extracting number of instances for each element in array
	for( int i = 0; i < input; i++)
	{
		count++;
		if(i==input-1 || arr[i] != arr[i+1]) {
			cout << left << setw(10) << arr[i] << left << setw(10) << count << endl;
			count =0;
			}
	}

[luv2program]

@Kevin C

yours works too and much more simpler O.O and thank you for the help. my head feels like its gonna explode.